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Let $f$ be meromorphic on $G$. Show that neither the poles nor the zeros of $f$ have a limit point in $G$.

Proof: 1. Let $Z=\{\text{zeros of } f\}$ and suppose that $Z$ has a limit point $b$ in $G$. Then exists $\{z_n\}\in Z$, $z_n$ are distinct and $z_n\neq b$ such that $z_n\to b$. Then $f(z_n)\to f(b)$ and hence $f(b)=0$. But we know that $f\equiv 0$ iff $\{z: f(z)=0\}$ has a limit point in $G$. Hence we get that $f\equiv 0$ but this is not the case since $f$ is meromorphic.

  1. Let $P=\{\text{poles of } f\}$ and suppose that $P$ has a limit point $a$ in $G$. Then exists $\{p_n\}\in P$, $p_n$ are distinct and $p_n\neq a$ such that $p_n\to a$. Since meromorphic function we can consider as continuous in $\mathbb{C}\cup \{\infty\}$ then $f(p_n)\to f(a)$ and hence $|f(a)|=\infty$. So $a$ is a pole of $f(z)$. Hence $f(z)(z-a)^k=g(z)$ for some $k\geq 1$ and $g$ is analytic function. If we plug in any point $p_n$ we get that $g(p_n)=\infty$ which is not the case.

Is my solution correct?

Would be very grateful for remarks!

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  • $\begingroup$ Not correct. Nothing prevents $g$ from having a pole at $p_n$. $\endgroup$ – user647486 Apr 14 '19 at 17:00
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    $\begingroup$ Analytic where? You haven't said. Certainly not on $G$. For example, if you take $f(z)=\frac{1}{(z-1)(z-2)}$ and $a=2$. Then $f(z)(z-2)^1=\frac{1}{z-1}=g(z)$ still has a pole at $z=1$. $\endgroup$ – user647486 Apr 14 '19 at 17:12
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    $\begingroup$ The proofs (1) and (2) are equivalent by working with $f$ and $1/f$ respectively and applying the identity principle. You can just deduce (2) by applying (1) to $1/f$. $\endgroup$ – user647486 Apr 14 '19 at 17:18
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    $\begingroup$ By the way, I hadn't noticed. The statement in the first sentence is not correct. Probably not properly quoted. The identity functions $f\equiv0$ and $f\equiv\infty$ are both meromorphic. Unless for some reason Conway deliberately excluded constant functions from his definition of meromorphic, which is not a common practice. $\endgroup$ – user647486 Apr 14 '19 at 17:23
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    $\begingroup$ Well, I can't. The statement is false as written. I checked his book and you didn't misquote him and his definition doesn't exclude constant functions. The statement should've been "Let $f$ be a non-constant meromorphic function on $G$ ...". For that statement, your proof of (1) is correct. $\endgroup$ – user647486 Apr 14 '19 at 17:28
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For the zeros you have a minor gap.

A meromorphic function $f$ on $G$ is a function that is holomorphic on all of $G$ except for a set $P$ of isolated points which are poles of $f$. The set $P$ is closed in $G$, hence $G' = G \setminus P$ is open in $G$ and therefore open in $\mathbb C$. Now $f$ is defined on all of $G'$. If the limit point $b$ of $Z$ is contained in $G'$, your agument works. So you have to show that $b \notin P$. But if $b \in P$, then $\lvert f(z)\rvert > 0$ for $z \in G$ with $\lvert z - b \rvert < \epsilon$ which is impossible since $\lvert z_n - b \rvert < \epsilon$ for $n \ge n_0$.

To deal with poles you can argue as follows. The function $g = \frac{1}{f}$ is defined on $G' \setminus Z$. From the first step we know that $Z$ is a set of isolated points. Moreover, since $f \ne 0$, all zeros must have finite order and we conclude that $g$ has a pole at each $z \in Z$. The singularities of $g$ at the points $p \in P$ are removable because $\lim_{z \to p} \lvert g(z) \rvert = \lim_{z \to p} \frac{1}{\lvert f(z) \rvert} = 0$. Hence $g$ extends to a homolomorphic function on $G \setminus Z$, in other words a meromorphic function on $G$ with zeros at all $p \in P$. Now apply again the first step.

The general picture in step 2 is this.

Let $\mathfrak{S}(G)$ denote the set of all pairs $(S,f)$ such that $S$ is a set of isolated points in $G$ and $f : G \setminus S \to \mathbb C$ is holomorphic function having either a removable singularity or a pole at each $s \in S$. Let $R(f)$, $P(f)$ and $Z(f)$ denote the set of removable singularities, poles and zeros of $f$, respectively. In $\mathfrak{S}(G)$ we can define an addition $(S,f) + (T,g) = (S \cup T,f + g)$, where $(f + g)(z) = f(z) + g(z)$ for $z \in G \setminus (S \cup T)$, similarly a multiplication $(S,f) \cdot (T,g)$. To simplify notation, we usually write $f$ instead of $(S,f)$.

The set $\mathfrak{M}(G)$ of meromorphic functions on $G$ is defined as the subset of $\mathfrak{S}(G)$ such that $R(f) = \emptyset$.

"Removing removable singularities" yields a surjective function $\rho : \mathfrak{S}(G) \to \mathfrak{M}(G)$.

The algebraic operations on $\mathfrak{S}(G)$ induce algebraic operations on $\mathfrak{M}(G)$ by $f + g = \rho(f + g)$, $f \cdot g = \rho(f \cdot g)$. It is easy to see that $\mathfrak{M}(G)$ with these operations is a field. The multiplicative inverse of $f \ne 0$ is $\rho(\frac{1}{f})$. In fact, $f$ is defined on $G \setminus P(f)$ and $\frac{1}{f}$ is defined on $G \setminus (P(f) \cup Z(f))$. The zeros of $f$ are poles of $\frac{1}{f}$ and the poles of $f$ are removable singularities of $\frac{1}{f}$ so that $\rho(\frac{1}{f})$ is defined on $G \setminus Z(f)$ with poles at the points of $Z(f)$ and zeros at the points of $P(f)$.

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